My favorite integer sequence

[jaydfox]

I'll also be much interested if I can learn about the approach how to determine the function f(x) which prognoses/approximates the elements of the sequence. I could implement the function f(x) in Pari/Gp but just did not get the clue how one would arrive at it.
Gottfried

Deriving the continuous function is actually pretty straightforward. I'll use exp(x) as an example:

\(
\exp(x) = \sum_{k=0}^{\infty} a_k x^k \\
a_0 = 1 \\
D\left[\exp(x)\right] = \exp(x)
\)

From this, you can derive exp(x).

\(
D\left[a_1 x\right] = a_0 \\
\Rightarrow a_1 = 1
\)

\(
D\left[a_2 x^2\right] = a_1 x \\
\Rightarrow 2 a_2 x = a_1 x \\
\Rightarrow a_2 = 1/2
\)

\(
D\left[a_3 x^3\right] = a_2 x^2 \\
\Rightarrow 3 a_3 x^2 = a_2 x^2 \\
\Rightarrow a_3 = \frac{1}{3}\left(\frac{1}{2}\right) = \frac{1}{6}
\)

In general:
\(
D\left[a_k x^k\right] = a_{k-1} x^{k-1} \\
\Rightarrow k a_k x^{k-1} = a_{k-1} x^{k-1} \\
\Rightarrow a_k = \frac{1}{k} \left(a_{k-1}\right)
\)

By induction:
\(
a_k = \frac{1}{k!}
\)


Now let's try with the problem at hand:

\(
f(x) = \sum_{k=0}^{\infty} b_k x^k \\
b_0 = 1 \\
D\left[f(x)\right] = f(x/2)
\)

From this, we start recursively solving:
\(
D\left[b_1 x\right] = b_0 \\
\Rightarrow b_1 = 1
\)

\(
D\left[b_2 x^2\right] = b_1 \left(\frac{x}{2}\right) \\
\Rightarrow 2 b_2 x = \frac{1}{2} b_1 x \\
\Rightarrow b_2 = \frac{1}{2} \left(\frac{1}{2}\right) = \frac{1}{4}
\)

\(
D\left[b_3 x^3\right] = b_2 \left(\frac{x}{2}\right)^2 \\
\Rightarrow 3 b_3 x^2 = \frac{1}{2^2} b_2 x^2 \\
\Rightarrow b_3 = \frac{1}{3} \left(\frac{1}{2^2}\right) \left(\frac{1}{4}\right) = \frac{1}{48}
\)

In general:
\(
D\left[b_k x^k\right] = b_{k-1} \left(\frac{x}{2}\right)^{k-1} \\
\Rightarrow k b_k x^{k-1} = \frac{1}{2^{k-1}} b_{k-1} x^{k-1} \\
\Rightarrow b_k = \frac{1}{k} \left(\frac{1}{2^{k-1}}\right) b_{k-1}
\)

By induction:
\(
b_k = \frac{1}{2^{(k-1)k/2} k!}
\)

Then, as I posted previously:

\(
f(x) = \sum_{k=0}^{\infty}\frac{1}{2^{k(k-1)/2} k!} x^k
\)